# Jonsson Boolean algebras?

Let us say that a mathematical structure of cardinality $\omega_1$ is Jonsson whenever every one of its proper substructures is countable.

There are examples of Jonsson groups due to Shelah or Obratzsov. I am almost sure that there is no Jonsson Boolean algebra but I cannot (dis)prove it by hand. Am I right?

PS. feel free to give any further examples of Jonsson structures or structures which are never Jonsson.

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My understanding is that it was Ol'shanskii who first constructed a countable Jonsson group (an infinite group all of whose subgroups are finite). Later Shelah constructed an uncountable Jonsson group (the so-called Kurosh monster). –  Ali Enayat Jul 11 '11 at 22:36
Since you asked for other examples: it is known that there are Jonsson models of PA (Peano Arithmetic). as well as ZFC (Zermelo-Fraenkel set theory with the axiom of choice) of power $\aleph_1$ in the following sense: there are models of $PA$ and $ZFC$ of power $\aleph_1$ that have no proper uncountable elementary submodel. This result is due to Julia Knight (Hanf numbers for omitting types over particular theories. J. Symbolic Logic 41 (1976), no. 3, 583–588). A different proof was given by Kossak and Schmerl in their book on models of $PA$. –  Ali Enayat Jul 11 '11 at 22:51
I can also recommend the following useful survey (alas, it does not seem to be available online). Coleman, Eoin; Jonsson groups, rings and algebras. Irish Math. Soc. Bull. No. 36 (1996), 34–45. The author's name appears is spelled OREN KOLMAN on his homepage. –  Ali Enayat Jul 12 '11 at 15:00

Boolean algebras are never Jonsson.

Suppose that $\mathbb{B}$ is a Boolean algebra of size $\omega_1$. Let $a$ be any element such that neither $a$ nor $\neg a$ is an atom. Note that every element $b\in\mathbb{B}$ is the join $b=(b\wedge a)\vee(b\wedge \neg a)$, and so there must be uncountably many elements either in the cone below $a$ or below $\neg a$. Assume without loss of generality that there are uncountably many elements below $a$. Let $\mathbb{C}$ be the subalgebra of $\mathbb{B}$ consisting of the elements below-or-equal $a$ or above-or-equal $\neg a$. This is closed under meets, joins and complements, and hence is a sub-Boolean algebra. And it has size $\omega_1$ by the choice of $a$. But it has no elements below $\neg a$ other than $0$, and so $\mathbb{C}$ is an uncountable proper subalgebra, as desired. QED

It seems that the same idea generalizes to any uncountable cardinal.

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I'm used to considering the theory of Boolean algebras as having two constants $0$ and $1$ as well (cf. Boolean rings); $\mathbb{C}$ is not a subalgebra in this sense. Do you know whether Boolean algebras in this sense can be Jonsson? –  Todd Trimble Jul 11 '11 at 18:00
But $0$ is below $a$ and $1$ is above $\neg a$, so they are in $\mathbb{C}$. Am I misunderstanding? –  Joel David Hamkins Jul 11 '11 at 18:02
My apologies! I misread your "or" as an "and". I'll have to rethink your example. –  Todd Trimble Jul 11 '11 at 18:07
That's interesting; thanks. +1. –  Todd Trimble Jul 11 '11 at 18:09
More generally, no abelian group of finite exponent (and hence no Boolean algebra) is Jonsson. To see this, simply note that any element is contained in a countable pure subgroup and pure subgroups of finite exponent are direct summands. So an uncountable such group will always have a non-trivial uncountable direct summand. –  Juris Steprans Jul 12 '11 at 1:57

Since Joel Hamkins has nicely answered the question about Boolean algebras, let me just present the following items dealing with the PS portion of the question.

(1) It is well-known that for any prime $p$, $\Bbb{Z}_{p^{\infty}}$ is a countable Jonsson group, and of course it is abelian; but constructing a countable non-abelian Jonsson group is much harder, and was accomplished by Ol'shanskii. There is more than one way to describe $\Bbb{Z}_{p^{\infty}}$. The quickest is: for a fixed prime $p$, $\Bbb{Z}_{p^{\infty}}$ is the collection of complex numbers that are the $p^n$-root of unity for some natural number $n$, equipped with complex multiplication.

(2) No uncountable abelian group is Jonsson (by the structure theorem for abelian groups).

(3) There are countable Jonsson fields in every characteristic; for characteristic $0$ this is clear since $\Bbb{Q}$ does the job, but for characteristic $p$ the fields are not widely known and are referred to as Steinitz fields; they are sometimes written as $GF(p^{q^{\infty}})$.

(4) No uncountable field is Jonsson. This follows from the fact that every uncountable field of cardinality $\kappa$ has a transcendence base of cardinality $\kappa$; which in turn implies that every field $F$ of uncountable power $\kappa$ has a subfield $F'$ of power $\kappa$ which is isomorphic to a purely transcendetal extension (of its prime field) of power $\kappa$, which of course has many ($2^\kappa$) subfields of power $\kappa$.

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On the Post scriptum and related to Boolean algebras, there are no Jónsson lattices of regular cardinality (T.P. Whaley, Large sublattices of a lattice, Pacific J. Math. 28 (1969), 477–484).

It is apparently still open whether there are no Jónsson lattices of singular cardinality (in ZFC).

A related question is whether there a non-trivial lattice that is not generated by the union of two proper sublattices, attributed to David Wasserman (Is there a nontrivial lattice that is not generated by the union of two proper sublattices?, manuscript, http://home.earthlink.net/~dwasserm/Sublattice.pdf) and discussed by George Bergman (Algebra univers. 55 (2006) 509–511), who notes that a Jónsson lattice would settle this.

There are no large Jónsson modules (over commutative rings) of regular or strong limit singular cardinality (where an R-module M is large if its cardinality is larger than that of R). See G. Oman, Some results on Jónsson modules over a commutative ring, Houston J. Math. 35 (2009), 1-12.

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Jónsson's 1972 book Topics in Universal Algebra has very accessible and still useful sections on the early results on Jónsson structures, and attributes the non-existence of Jónsson Boolean Algebras to Tarski, if I recall correctly. –  Oren Kolman Sep 5 '11 at 14:31

A quick proof that no uncountable abelian group is Jonsson goes like this: Suppose $G$ is such a group. Then $G$ is either divisible or has a maximal subgroup $M$. If a maximal subgroup $M$ exists, then $G/M$ is of order $p$, whence $|M|=|G|$, and $G$ is not Jonsson. Thus $G$ is divisible, and hence is a direct sum of copies of $\mathbb{Q}$ and $C(p^\infty)$ for various primes $p$ (all such groups are countably infinite). Simply delete one of the summands, and you get a proper subgroup of G of the same cardinality as G, and we have reached a contradiction.

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